Showing $\Gamma = \{(x,y) \in \mathbb{R}^d \times \mathbb{R}: y = f(x)\}$ is measurablle and of measure 0, when $f$ is a measurable function. 
Exercise 2.7 (Stein): If $\Gamma \subset \mathbb{R}^d \times \mathbb{R},$ $\Gamma = \{(x,y) \in \mathbb{R}^d \times \mathbb{R}: y = f(x)\}$, and assume $f$ is measurable on $\mathbb{R}^d$. Show that $\Gamma$ is a measurable subset of $\mathbb{R}^{d+1}$ and $m(\Gamma) = 0$.

Where my intuition leads me:
I know how to show $m(\Gamma) = 0$ if $\Gamma$ is measurable (I think). Essentially we need to slice (and dice, I haven't eaten yet) up $\Gamma$ into a bunch of smaller sets for arbitrary fixed $x \in \mathbb{R}^d$:
\begin{equation*}
\Gamma_x := \{y \in \mathbb{R}: (x,y) \in \Gamma\} = \{f(x)\}.
\end{equation*}
It's clear that the measure of a singleton is zero, and thus
\begin{equation*}
m(\Gamma) = \int_{\mathbb{R}^d}m(\Gamma_x)dx = 0.
\end{equation*}
Remark: This is ensured from a previous Corollary that says we can do this for almost every $x \in \mathbb{R}^d \times \mathbb{R}$ given that $\Gamma \subset \mathbb{R}^d \times \mathbb{R}$ is measurable.
Now, I'm kind of stuck on proving that $\Gamma$ is measurable. I'm thinking that I should use some variant of the following Corollary

Corollary 2.3.8: Suppose $f(x)$ is a non-negative function on $\mathbb{R}^d$, and let
\begin{equation*}
A = \{(x,y) \in \mathbb{R}^d \times \mathbb{R}: 0 \leq y \leq f(x)\}
\end{equation*}
Then $f$ is measurable on $\mathbb{R}^d$ IFF $A$ is measurable in $\mathbb{R}^{d+1}$.

I imagine I'd have to formulate some sequence
\begin{equation*}
E_n := \biggr\{(x,y) \in \mathbb{R}^d \times \mathbb{R}: f(x) - \frac{1}{n} \leq y \leq f(x)\biggr\},
\end{equation*}
which converges to $\Gamma$, and prove each member of this sequence is measurable. My idea is that since $f$ is measurable on $\mathbb{R}^d$ (suffices to assume that it is non-negative), then we know that
\begin{equation*}
F(x,y) = y - f(x).
\end{equation*}
Now, $F$ is measurable on $\mathbb{R}^{d+1}$, and $E_n = \{y \geq f(x) - \frac{1}{n}\} \cap \{F \leq 0\}$. Measurability follows.
Is this the right track for proving measurability of $\Gamma$?
I feel a little shaky on the introduction of $F(x,y)$ as being a valid procedure (I'm basing this off of the proof of the Corollary I'm trying to alter)
 A: To show the measurability of the set $\Gamma$, it suffices that both $\left\{ (x,y)\mid y>f(x)\right\} $
and $\{(x,y)\mid y<f(x)\}$ are measurable. Observe that
\begin{eqnarray*}
 &  & \{(x,y)\in\mathbb{R}^{d}\times\mathbb{R}\mid y>f(x)\}\\
 & = & \cup_{r\in\mathbb{Q}}\{(x,y)\mid y>r>f(x)\}\\
 & = & \cup_{r\in\mathbb{Q}}\{(x,y)\mid y>r\}\cap\{(x,y)\mid r>f(x)\}\\
 & = & \cup_{r\in\mathbb{Q}}[\mathbb{R}^{d}\times(r,\infty)]\cap[f^{-1}((-\infty,r))\times\mathbb{R}]
\end{eqnarray*}
Now, it is clear that $\{(x,y)\in\mathbb{R}^{d}\times\mathbb{R}\mid y>f(x)\}\in\mathcal{B}(\mathbb{R}^{d}\times\mathbb{R})=\mathcal{B}(\mathbb{R}^{d})\otimes\mathcal{B}(\mathbb{R})$.
The measurability of $\{(x,y)\mid y<f(x)\}$ can be proved similarly.

To prove that $m(\Gamma)=0$, we may invoke Fubini-Tonelli Theorem.
The Lebesgue measure $\lambda$ on $\mathbb{R}^{d}\times\mathbb{R}$
is $\lambda=m_{d}\times m$, where $m_{d}$ and $m$ are Lebesgue
measures on $\mathbb{R}^{d}$ and $\mathbb{R}$ respectively. We have
that
\begin{eqnarray*}
\lambda(\Gamma) & = & \int1_{\Gamma}(x,y)d\lambda(x,y)\\
 & = & \int\left(\int1_{\Gamma}(x,y)dm(y)\right)dm_{d}(x).
\end{eqnarray*}
For each $x\in\mathbb{R}^{d}$, $1_{\Gamma}(x,y)=1$ iff $y=f(x)$.
Therefore $\int1_{\Gamma}(x,y)dm(y)=m(\{f(x)\})=0$ (The measure of
a singleton is zero!). It follows that $\lambda(\Gamma)=0$.
A: Here is another proof of the measurability of $E=\{(x,y):f(x)=y\}$.
The map $G:(x,y)\mapsto(f(x),y)$ from $X\times Y$ into $Y\times Y$ is $\mathcal{B}(\mathbb{R}^n)\otimes\mathcal{B}(\mathbb{R})-(\mathcal{B}(\mathbb{R})\otimes\mathbb{R}(\mathcal{B}(\mathbb{R}))$ measurable
since $G^{-1}(A\times B)=f^{-1}(A)\times B$ for all $A\times B\subset \mathbb{R}\times\mathbb{R}$.
The map $S:(u,v)\mapsto u-v$ from $\mathbb{R}\times\mathbb{R}$ into $\mathbb{R}$ is $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})-\mathcal{B}(\mathbb{R})$ measurable since $S$ is continuous and by separability of $\mathbb{R}$, $\mathcal{B}(\mathbb{R}\times\mathbb{R})=\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$.
Hence $(S\circ G)(x,y)=f(x)-y$ is $\mathcal{B}(\mathbb{R}^n)\otimes\mathcal{B}(\mathbb{R})-\mathcal{B}(\mathbb{R})$ measurable.
Thus  $E=\{(x,y):f(x)=y\}=(S\circ G)^{-1}(\{0\})$ is a measurable set in $\mathcal{B}(\mathbb{R}^n)\otimes\mathcal{B}(\mathbb{R})$.

The rest of the problem can be dealt with with Fubini's theorem, as it I suggested (implicitly) in the OP.
